how to do bitxor operation of two 1*255 matrix

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moni sinha
moni sinha 2018 年 11 月 30 日
編集済み: Greg 2018 年 11 月 30 日
h1 =1x255 logical
h3 = 1x255 logical
howto do bitxor of h1 and h3

回答 (2 件)

Greg
Greg 2018 年 11 月 30 日
編集済み: Greg 2018 年 11 月 30 日
result = h1 | h3;
Edit: this is logical (bit) or, not xor. As posted elsewhere, simply use the xor function.
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Greg
Greg 2018 年 11 月 30 日
Good catch Guillaume, i kept reading or not xor.
Why are we assuming c is the second argument? The original post explicitly states h1 and h3, both are logical and same size. All following posts are new problems to the original question.

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James Tursa
James Tursa 2018 年 11 月 30 日
編集済み: James Tursa 2018 年 11 月 30 日
It's not entirely clear to me what operation you really want, but if the elements of h1 and h2 represent "bits", then you could just do this:
result = (h1 ~= h2); % equivalent of xor between the elements of h1 and h2
If h1 and h2 don't have the same number of elements, then that is a different problem that you will need to fix before doing the xor operation.
  2 件のコメント
Greg
Greg 2018 年 11 月 30 日
Guillaume's comment here should be a separate answer, and accepted.

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